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Three dwarves and three elves sit down in in the row of six [#permalink]

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19 Jan 2006, 12:06

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Three dwarves and three elves sit down in in the row of six chairs. If no dwarf will sit next to another dwarf and no elf will sit next to another elf, in how many different ways can the elves and dwarves sit.

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19 Jan 2006, 17:40

The elves and dwarves must alternate:

e d e d e d or d e d e d e

Considering the first case, the total number ways is to rotate the elves 3! times while keeping the dwarves to one combination. Then rotate the dwarves 3! times while keeping the elves to one combination.

However, that is only for the first case. We must multiply by 2 as we need to account for the second case.

Re: Three dwarves and three elves sit down in in the row of six [#permalink]

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07 Jul 2013, 15:31

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Three dwarves and three elves sit down in in the row of six chairs. If no dwarf will sit next to another dwarf and no elf will sit next to another elf, in how many different ways can the elves and dwarves sit.

My approach (2 mins 20 seconds):

We have 6 slots. The first slot could be filled with any of the 6 dwarves or elves (then 6 possibilities). The second, could only be filled with the other half of the group, depending if the first was filled with a dwarve or an elve (then 3 possibilities). The third, could only be filled with the 2 dwarves or the 2 elves remaining (then 2 possibilities). And the fourth could only be filled with the other 2 dwarves or elves (2 possibilities). The rest is filled with either one elve or one dwarve.

6*3*2*2= 72
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Three dwarves and three elves sit down in in the row of six chairs. If no dwarf will sit next to another dwarf and no elf will sit next to another elf, in how many different ways can the elves and dwarves sit.

My approach (2 mins 20 seconds):

We have 6 slots. The first slot could be filled with any of the 6 dwarves or elves (then 6 possibilities). The second, could only be filled with the other half of the group, depending if the first was filled with a dwarve or an elve (then 3 possibilities). The third, could only be filled with the 2 dwarves or the 2 elves remaining (then 2 possibilities). And the fourth could only be filled with the other 2 dwarves or elves (2 possibilities). The rest is filled with either one elve or one dwarve.

6*3*2*2= 72

Three dwarves and three elves sit down in a row of six chairs. If no dwarf will sit next to another dwarf and no elf wil sit next to another elf, in how many different ways can the elves and dwarves sit?

In order to meet the restriction dwarves and elves must sit either DEDEDE or EDEDED. There are 3!*3!=36 arrangements possible for each case (3! arrangements of dwarves and 3! arrangements of elves), so total ways to sit are 2*36=72.

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Does the formula 6!/3!3! only work if they are sitting next to each other?

e.g. sitting DDDEEE or EEEDDD.

Or is it due to having 2 different sub-sets that we can't use such formula.

Thank you for the help.

6!/(3!3!)=20 is the number of arrangements of 6 letters EEEDDD, where 3 D's and 3 E's are identical:EEEDDD;EEDEDD;EDEEDD;DEEEDD;...DDDEEE.

But in our original question we don't have 3 identical D's and 3 identical E's. Also, we need only those arrangements where no dwarf will sit next to another dwarf and no elf will sit next to another elf.